h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

Galileo discovered that irrespective of the mass of an object being dropped, objects fall at the same rate. A grand piano dropped from a tall building will fall at the same rate as a golf ball (1).

So we may deduce from this that mass is does not matter when it comes to the acceleration of an object due to gravity.

However, we see that mass does matter, because acceleration due to gravity is greater on earth than it is on the moon because the earth has a greater mass than the moon and therefore greater gravitational pull. (2)

So which is correct? An object's mass effects the rate of acceleration due to gravity (2) or it doesn't (2)?

Galileo discovered that irrespective of the mass of an object being dropped, objects fall at the same rate. A grand piano dropped from a tall building will fall at the same rate as a golf ball (1).

So we may deduce from this that mass does not matter when it comes to the acceleration of an object due to gravity.

However, we see that mass does matter, because acceleration due to gravity is greater on earth than it is on the moon because the earth has a greater mass than the moon and therefore greater gravitational pull. (2)

So which is correct? An object's mass effects the rate of acceleration due to gravity (2) or it doesn't (1)?

Somebody with a far better understanding than me will be along soon I expect, but I would have thought the strength of gravity merely affects the rate of acceleration to terminal velocity, which should be a constant.

But I'm probably wrong.

Almost certainly, in fact,

Gravity is far too heavy a subject to think about at 1am.

F = ma

Force = mass x acceleration.

Fgrav = m1 x m2 x G /d^2

The force between two objects is the product of the masses of those objects, times G and divided by the square of the distance between them.

Where m1 is the mass of the earth and m2 is the mass of something falling towards it...

So G is 6.67 x 10-11 N m2/kg2. Or put another way, a constant.
And d^2 is something small, unless you're falling from a LONG way.
And m1, the mass of the earth, is 6 × 10^24 kg.
So unless m2 is ****ing HUGE, it kind of doesn't matter how big it is.
If it's 1 kg or 100 kg or 10,000 kg... in the scheme of things, it's nothing.

OK?

SoRB

What SoRB says it true if both objects are coming together in a vacuum - in such a scenario there is no terminal velocity except in as much as only at the moment before collision will a maximal velocity be achieved.

Terminal velocity is only reached if there is some friction opposing the acceleraton towards the centre of the earth. The object will accelerate until the opposing force matches the gravitational force. The atmosphere does pretty well at this.
So a feather dropped from the Leaning Tower of Pisa (or any other tall place of your choosing) will fall more slowly than grand piano because of the friction caused by the air opposing the gravitational pull. We have cunningly designed a device to make use of this effect. It's called a parachute

I can't remember who it was (someone later than Gallileo) famously put a penny and a feather in a vacuum tube and showed that under these conditions they do indeed fall at the same rate. A strong piece of supporting evidence for the law of gravity posted by SoRB above.

Thought I'd better cross-reference this to the "Whatever it is we're talking about now thread", since it cropped up there, too: http://www.bbc.co.uk/dna/h2g2/brunel/F19585?thread=4005961&skip=2060&show=20 <> Maybe not. I'm not a physicist, but I've been wondering about gravity and been asking questions to try to get my head around something. I asked a question on this forum a while ago - a thread which I can't find at the moment. Essentially my question was "How come the acceleration due to gravity isn't affected by the mass of the object dropped, but it is affected by the mass of the planet/moon/big_thing?" Obviously, observation shows that to all intents and purposes objects of differing masses fall at the same rate. But I just wondered why the mass of the small object has no bearing at all, when that of the large object clearly does. What if we are dealing not with one huge mass and one relatively tiny mass. What if we are dealing with two roughly similar-sized masses. Anyway, I think that's what the answers were leading to. The force of gravity on an object F = mg where m is the mass of the object and g is acceleration. We are interested here in g. Acceleration due to gravity is expressed thus: g = F/m Now, F = GMm/R2 so g = (GMm/R2)/m In this equation the rate of acceleration will always be the same because the mass of the smaller object cancels itself out, but this may be an oversimplification.

<>

In the case of something falling from a tree on earth, d is about 4000 miles (the approx radius of the earth) plus the height of the tree?

F = m x a

Presumably the earth (m1) hits an object (m2) with the same force as the object hits the earth.

So:
m1 x a1 = m2 x a2
Therefore
F = (m1.a1 + m2.a2)/2

If it's also true that:
F = m1.m2.G/d^2
then:
(m1.a1 + m2.a2)/2 = m1.m2.G/d^2

...which gives us this result:

m2.a2 + m1.a1 = 2m1.m2.G/d^2

So
m2.a2 = 2m1.m2.G/d^2 - m1.a1

Let's divide both sides by m2:

a2 = 2m1.G/d^2 - m1.a1/m2

Unless I've made a mistake somewhere, which is not inconceivable, this:
a2 = 2m1.G/d^2 - m1.a1/m2

tells us that a small object will accelerate less quickly than a more massive one.

In which case Gallileo was wrong

Sure gravity doesn't exist anyways - it's just intelligent falling.

It's a case of things not mattering at this scale. There is no discernable difference between small masses (on Earth, anything you can conceive of falling counts as a small mass in relation to the Earth).

TRiG.

Well, I guess it's the a1 bit (acceleration of earth), which must be almost zero, that renders the difference in a2 insignificant.

a2 = 2m1.G/d^2 - m1.a1/m2

When I tried putting figures in, I got the result a2 = 20m/s/s

This is about twice what it should be. I think I might have made a mistake in post 8. Since G is the force between the two masses, there's no need to double it.

I'll try again, _properly_ this time:

F = m x a

Presumably the earth (m1) hits an object (m2) with the same force as the object hits the earth.

So:
m1 x a1 = m2 x a2
Therefore
F = m1.a1 + m2.a2

If it's also true that:
F = m1.m2.G/d^2
then:
m1.a1 + m2.a2 = m1.m2.G/d^2



So
m2.a2 = m1.m2.G/d^2 - m1.a1

Let's divide both sides by m2:

a2 = m1.G/d^2 - m1.a1/m2

Let's check.
The earth (m1) is about 6,000,000,000,000,000,000,000,000kg in mass.
Its radius is about 6,400,000m.

a2 = m1.G/d^2 - m1.a1/m2

= (6x10^24kg x G) / d^2 - m1.a1/m2

= (6x10^24kg x 6.67x10^-11Nm^2/kg^2)/d^2 - m1.a1/m2

= (6x10^24 x 6.67x10^-11Nm^2/kg)/d^2 - m1.a1/m2

= (6 x 6.67x10^13Nm^2/kg)/d^2 - m1.a1/m2

= (40x10^13Nm^2/kg)/(6400000m)^2 - m1.a1/m2

= (40x10^13Nm^2/kg)/41 x 10^12m^2 - m1.a1/m2

= (40x10Nm^2/kg)/41m^2 - m1.a1/m2

= (400N/kg)/41 - m1.a1/m2

= (400N/kg)/41 - m1.a1/m2

= 9.8 m/s/s - m1.a1/m2


Seems OK.

OUCH!



Mind where you're putting those numbers, if you please!

Hi WG,

I think you do need that 2 in there, actually, since F=m1.a1 therefore m1.a1+m2.a2 = 2F

Damned if I can see any error in this, looks like heavy objects do fall faster!

In addition to this, the same force acts on the Earth, so heavy objects will also accelerate the Earth towards them and thus hit the Earth sooner. (Nobody mention Einstein!)

Gif

<>

That's what I thought at first.
But we have to be careful about what we mean by F.

In post 4 Sorb said the force Fgrav BETWEEN two objects is the product of the masses of those objects, times G and divided by the square of the distance between them.

I think "between" is the key word here, and the reason we don't need the 2.

The ~force Fgrav between the two objects~

must equal

~the sum of
F1 (the mass m1 of the earth times its acceleration) and F2 (the mass m2 of the apple (ball, etc) times its acceleration)~

So Fgrav = F1 + F2

= m1.a1 + m2.a2